
\prob{00A5}{分部积分法}

证明：对于有连续导数$u', v'$的函数$u, v$，有
\[ \int u \dif v = uv - \int v \dif u \]
\problabels{yellow/微积分, green/证明题}

\subsection{函数相乘求导}

由
\[ \left(u(x)v(x)\right)' = u'(x)v(x) + u(x)v'(x) \]
移项得
\[ u(x)v'(x) = \left(u(x)v(x)\right)' - u'(x)v(x) \]
对两边求不定积分，得
\[ \int u(x)v'(x) \dif x = u(x)v(x) - \int u'(x)v(x) \dif x \]
将$u'(x)$写作$\dif u/\dif x$，$v'(x)$写作$\dif v/\dif x$，于是上式可简化为
\[ \int u \dif v = uv - \int v \dif u \]
证毕。
